Geometric proofs with vectors: parallel, perpendicular, midpoint and ratio properties

Syllabus dot point

“Use vector methods to prove geometric properties, including parallelism, perpendicularity, midpoint and ratio division”

HSCMaths Extension 1Vectors (ME-V1)14 min readUpdated 20 Jun 2026

Quick answer

Vector proofs run a fixed four-step tactic: assign position vectors, express displacements with $\overrightarrow{XY} = \mathbf{y} - \mathbf{x}$ , compute the algebraic relation, and read off the geometry. Parallel means a scalar multiple, perpendicular means a zero dot product, a midpoint is the average $\frac{\mathbf{a}+\mathbf{b}}{2}$ , and the section formula $\frac{n\mathbf{a}+m\mathbf{b}}{m+n}$ divides $AB$ in the ratio $m:n$ . From these, results like the midpoint connector theorem and the diagonal-bisection property follow in a few lines.

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

NESA wants you to prove geometric facts using the algebra of vectors instead of coordinate geometry or congruent-triangle arguments: that two lines are parallel, perpendicular or equal in length, that a point is a midpoint, that a segment is divided in a given ratio, or that a figure is a parallelogram, rhombus or rectangle. The appeal is that one or two lines of vector algebra often replace a page of coordinate work, and the result holds for any configuration, not just the one in the diagram.

The answer

Position vectors and the displacement rule

Choose an origin $O$ . Each point $P$ then has a position vector $\mathbf{p} = \overrightarrow{OP}$ . The single rule that drives every proof is the displacement formula: the vector from one point to another is the difference of their position vectors.

For points $A$ , $B$ , $C$ with position vectors $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ ,

\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, \qquad \overrightarrow{AC} = \mathbf{c} - \mathbf{a}.

Everything below is built from this one identity.

The four-step tactic (use this every time)

Vector proofs all follow the same shape. Treat it as a fixed procedure:

Assign a position vector to every labelled point (and, if it helps, put the origin at a useful vertex so one or more vectors become $\mathbf{0}$ ).
Express the displacement vectors the question is about, using $\overrightarrow{XY} = \mathbf{y} - \mathbf{x}$ .
Compute the algebraic relation that encodes the geometric claim, an equality, a scalar multiple, or a zero dot product.
Conclude by translating that relation back into the geometric statement (and say it in words).

The skill is matching the geometric claim to the right algebraic relation in step 3. Here is the dictionary.

Key fact

Translate the geometry into algebra:

Parallel ( $\overrightarrow{XY} \parallel \overrightarrow{UV}$ ): one is a scalar multiple of the other, $\overrightarrow{XY} = \lambda\,\overrightarrow{UV}$ .
Equal and parallel ( $\overrightarrow{XY} = \overrightarrow{UV}$ ): same direction and length, the test for a parallelogram.
Collinear ( $X$ , $Y$ , $Z$ on a line): $\overrightarrow{XY} = \lambda\,\overrightarrow{XZ}$ (parallel sharing a point).
Perpendicular ( $\overrightarrow{XY} \perp \overrightarrow{UV}$ ): $\overrightarrow{XY} \cdot \overrightarrow{UV} = 0$ .
Equal length ( $|XY| = |UV|$ ): $|\overrightarrow{XY}|^2 = |\overrightarrow{UV}|^2$ , expanded with the dot product.
Midpoint $M$ of $AB$ : $\mathbf{m} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Midpoint and section formulas

The midpoint $M$ of $AB$ is the average of the endpoints:

\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}.

More generally, the point $P$ dividing $AB$ internally in the ratio $m : n$ (so $\dfrac{AP}{PB} = \dfrac{m}{n}$ ) has position vector

\mathbf{p} = \frac{n\,\mathbf{a} + m\,\mathbf{b}}{m + n}.

The cross-over of the labels is worth memorising: the point nearer $B$ (large $m$ ) carries the weight $m$ on $\mathbf{b}$ . The midpoint is just the case $m : n = 1 : 1$ . (For external division, beyond $B$ , replace $n$ by $-n$ , taking care with signs.)

Parallelogram and rectangle criteria

Parallelogram: $ABCD$ is a parallelogram iff $\overrightarrow{AB} = \overrightarrow{DC}$ (one pair of opposite sides equal and parallel). Equivalently, the diagonals bisect each other: the midpoint of $AC$ equals the midpoint of $BD$ .
Rhombus: a parallelogram with two equal adjacent sides, $|\overrightarrow{AB}| = |\overrightarrow{AD}|$ .
Rectangle: a parallelogram with a right angle, $\overrightarrow{AB} \cdot \overrightarrow{AD} = 0$ .

Be careful with vertex order in $\overrightarrow{AB} = \overrightarrow{DC}$ : it is $\overrightarrow{DC}$ , not $\overrightarrow{CD}$ , because going $A \to B$ must match going $D \to C$ for the figure to close as $ABCD$ .

Two results the proofs below establish

The midpoint connector theorem says the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

A second classic: the diagonals of a parallelogram bisect each other, meeting at their common midpoint $M$ .

How exam questions ask about vector proofs

"Prove that $PQRS$ is a parallelogram": show one pair of opposite sides are equal vectors, $\overrightarrow{PQ} = \overrightarrow{SR}$ .
"Show that $XY$ is parallel to $UV$ ": show $\overrightarrow{XY}$ is a scalar multiple of $\overrightarrow{UV}$ (and state the ratio of lengths if asked).
"Prove the diagonals bisect each other": show the midpoints of the two diagonals have the same position vector.
"Prove that the figure is a rhombus / rectangle": for a rhombus, equal adjacent side lengths; for a rectangle, a zero dot product of adjacent sides.
" $P$ divides $AB$ in the ratio $m : n$ ; find its position vector": apply the section formula $\frac{n\mathbf{a} + m\mathbf{b}}{m + n}$ .
"Show that $X$ , $Y$ , $Z$ are collinear": show $\overrightarrow{XY} = \lambda\,\overrightarrow{XZ}$ .

Exam technique

Always define your position vectors in the first line ("Let $\overrightarrow{OA} = \mathbf{a}$ , ..."); markers award a mark just for setting up the vectors cleanly. Choosing the origin at a vertex (so that vertex's position vector is $\mathbf{0}$ ) often halves the algebra, do it when you can. State the geometric conclusion in words after the algebra ("so $\overrightarrow{PQ} = \overrightarrow{SR}$ , hence $PQRS$ is a parallelogram"), because the final mark is usually for the interpretation, not the computation. Keep the proof origin-independent: do not assign coordinates to points whose coordinates are not given. For a "parallel and half the length" conclusion, the scalar multiple $\frac{1}{2}$ gives both facts at once, point that out.

Worked example

Prove the midpoint connector theorem

Assign position vectors and express the displacement: Triangle $ABC$ has position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ for $A, B, C$ . Let $Q$ and $R$ be the midpoints of $CA$ and $AB$ . By the midpoint formula,

Q = \frac{\mathbf{c} + \mathbf{a}}{2}, \qquad R = \frac{\mathbf{a} + \mathbf{b}}{2}.

Express the connector $\overrightarrow{RQ}$ by subtracting position vectors:

\overrightarrow{RQ} = Q - R = \frac{\mathbf{c} + \mathbf{a}}{2} - \frac{\mathbf{a} + \mathbf{b}}{2} = \frac{\mathbf{c} - \mathbf{b}}{2}.

Compute the relation and conclude: Since $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$ , we have $\overrightarrow{RQ} = \tfrac{1}{2}\,\overrightarrow{BC}$ . The scalar multiple $\frac{1}{2}$ says it all: $RQ$ is parallel to $BC$ (a scalar multiple) and half its length (the scalar is $\frac{1}{2}$ ). That is the midpoint connector theorem.

Prove the diagonals of a parallelogram bisect each other

In parallelogram $ABCD$ , opposite sides are equal and parallel, so $\overrightarrow{AB} = \overrightarrow{DC}$ , that is $\mathbf{b} - \mathbf{a} = \mathbf{c} - \mathbf{d}$ , which rearranges to

\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}.

The midpoint of diagonal $AC$ is $\frac{\mathbf{a} + \mathbf{c}}{2}$ and the midpoint of diagonal $BD$ is $\frac{\mathbf{b} + \mathbf{d}}{2}$ . Because $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$ , these are equal, so both diagonals pass through the same midpoint: they bisect each other.

Apply the section formula

$P$ divides $AB$ internally in the ratio $1 : 3$ , meaning $AP : PB = 1 : 3$ . With $m : n = 1 : 3$ (the ratio measured from $A$ ), the section formula gives

\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m + n} = \frac{3\mathbf{a} + 1\cdot\mathbf{b}}{1 + 3} = \frac{3\mathbf{a} + \mathbf{b}}{4}.

The weight on $\mathbf{a}$ is larger because $P$ is closer to $A$ .

Derive the cosine rule by expanding a squared magnitude

Let two sides of a triangle from one vertex be $\mathbf{a}$ and $\mathbf{b}$ with included angle $\theta$ , so the third side is $\mathbf{a} - \mathbf{b}$ . Expand its squared length with the dot product, using $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ :

|\mathbf{a} - \mathbf{b}|^2 = (\mathbf{a} - \mathbf{b})\cdot(\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - 2\,\mathbf{a}\cdot\mathbf{b} + |\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta.

This is the cosine rule, derived in three lines of vector algebra.

Prove perpendicular diagonals force a rhombus

Let $ABCD$ be a parallelogram with $\overrightarrow{AB} = \mathbf{u}$ and $\overrightarrow{AD} = \mathbf{v}$ . Its diagonals are $\overrightarrow{AC} = \mathbf{u} + \mathbf{v}$ and $\overrightarrow{DB} = \mathbf{u} - \mathbf{v}$ . Suppose the diagonals are perpendicular, so their dot product is zero:

(\mathbf{u} + \mathbf{v})\cdot(\mathbf{u} - \mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2 = 0 \quad\Longrightarrow\quad |\mathbf{u}| = |\mathbf{v}|.

So adjacent sides $AB$ and $AD$ are equal in length, and a parallelogram with equal adjacent sides is a rhombus.

Common traps

Assuming coordinates that are not given: A vector proof must hold for any origin and any configuration. Use position vectors, not made-up coordinates, unless the question supplies them.
Confusing "parallel" with "equal": Parallel means scalar multiple; equal means same direction and length. A parallelogram needs $\overrightarrow{AB} = \overrightarrow{DC}$ , not merely parallel sides.
Vertex order in the parallelogram condition: It is $\overrightarrow{AB} = \overrightarrow{DC}$ (so the figure closes as $ABCD$ ), not $\overrightarrow{AB} = \overrightarrow{CD}$ . Getting the order wrong reverses a side.
Section-formula labels swapped: For ratio $m : n$ from $A$ , the position vector is $\frac{n\mathbf{a} + m\mathbf{b}}{m + n}$ : the weights cross over. The midpoint is the special case $1 : 1$ .
Skipping the dot-product test for perpendicularity: "It looks like a right angle" is not a proof. Show $\mathbf{u}\cdot\mathbf{v} = 0$ .
Stopping at the algebra: After deriving the relation, state the geometric conclusion in words. The final mark is for the interpretation.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style5 marks

ABCD

is a quadrilateral. Let

P

Q

R

S

be the midpoints of

AB

BC

CD

DA

respectively. Use vectors to prove that

PQRS

is a parallelogram.

Show worked answer →

Let the position vectors of $A$ , $B$ , $C$ , $D$ be $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ , $\mathbf{d}$ .

Midpoints: $P = \frac{\mathbf{a} + \mathbf{b}}{2}$ , $Q = \frac{\mathbf{b} + \mathbf{c}}{2}$ , $R = \frac{\mathbf{c} + \mathbf{d}}{2}$ , $S = \frac{\mathbf{d} + \mathbf{a}}{2}$ .

Vector $\mathbf{PQ} = Q - P = \frac{\mathbf{b} + \mathbf{c}}{2} - \frac{\mathbf{a} + \mathbf{b}}{2} = \frac{\mathbf{c} - \mathbf{a}}{2}$ .

Vector $\mathbf{SR} = R - S = \frac{\mathbf{c} + \mathbf{d}}{2} - \frac{\mathbf{d} + \mathbf{a}}{2} = \frac{\mathbf{c} - \mathbf{a}}{2}$ .

So $\mathbf{PQ} = \mathbf{SR}$ , which means $PQ$ is parallel to $SR$ and equal in length.

Therefore $PQRS$ is a parallelogram.

Markers reward labelling position vectors of $A, B, C, D$ , computing each midpoint, computing both sides of the candidate parallelogram, and observing equality.

HSC-style4 marks

OACB

is a parallelogram with

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OB} = \mathbf{b}

. Prove that the diagonals

OC

and

AB

bisect each other.

Show worked answer →

Since $OACB$ is a parallelogram, $\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$ .

Midpoint of $OC$ has position vector $\frac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Midpoint of $AB$ has position vector $\frac{1}{2}(\mathbf{a} + \mathbf{b})$ (the average of $A$ at $\mathbf{a}$ and $B$ at $\mathbf{b}$ ).

The two midpoints are equal, so the diagonals meet at their common midpoint, that is they bisect each other.

Markers reward expressing the diagonal as $\mathbf{a} + \mathbf{b}$ , computing both midpoints, and concluding from their equality.

Practise this

Sources & how we know this

Mathematics Extension 1 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 1 HSC exam papers — NESA